Test de primitives amstex

Part I
Some tests

1 Matrices

Toutes les références

Les numéros sont :

1.b3.a3.b3.c

1.1 Simple

10
01
  
⎛
⎜
⎝
10
01
⎞
⎟
⎠
  
⎪
⎪
⎪
⎪		⎪
⎪
⎪
⎪
10
01
12
⎪
⎪
⎪
⎪		⎪
⎪
⎪
⎪
(1.a)

1.2 Compliqué

⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
1…………………………………………………
…2 ………………………………………………
……3 ……………………………………………
………4 …………………………………………
…………5 ………………………………………
……………6 ……………………………………
………………7 …………………………………
…………………8 ………………………………
……………………9 ……………………………
………………………10 …………………………
…………………………11 ………………………
……………………………12 ……………………
………………………………13 …………………
…………………………………14 ………………
……………………………………15 ……………
………………………………………16 …………
…………………………………………17 ………
……………………………………………18 ……
………………………………………………19 …
…………………………………………………20
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
 (1.b)

2 Environement cases

To summarize, we obtain the following exact representation for the inverse of x2ex+1:

⎧
⎪
⎪
⎨
⎪
⎪
⎩
Y(x) = y(log x), y0 inverse of 2log x+x+log(1+e−x/x2),
y[x+log(1+y0−2(x)e−y0(x))] = y0(x), y0 inverse of x+2log x,
y0(x) = y1(log x), y1 inverse of log x+log(1+2log x/x),
y1(x) = exp(y2(x)), y2 inverse of x+log(1+2xe−x),
y2[x+log(1+2y3(x)e−y3(x))] = y3(x),y3 inverse of x.

3 Test d’alignement

Now the φis are very easy to compute:

φ1 = y0 = 1/t2,         
φ2
= φ1(y0(x+g))−φ1 =
t3t22
(1+2t2)
 −
1+4t2+2t22
2(1+2t2)3
t24t32+O(t33),
         
φ3
= φ2(y0(x+g))−φ2 = −
1+4t2+2t22
(1+2t2)3
t24t32+O(t33),
         

A similar treatment applies to (1.b), and leads to

x+log(1+2xe−x) = 1/t1(y3(x+g))    = x+2xe−x−2x2e−2x+O(x3e−3x),       
exp[−x+log(1+2xe−x)] = t2(y3(x+g))    = e−x−2xe−2x+4x2e−3x+O(x3e−4x).        

3.1 gather, multline

or one of the following (successive) refinements:

  exp(eU)⎡
⎢
⎢
⎣		1−
2e−U1/2
U1/2+4
 +
2e−2U1/2
(U1/2+4)2
 −
4
3
e−3U1/2
(U1/2+4)3
+O(e−4U1/2)⎤
⎥
⎥
⎦		,
(3.a)
exp(eU)exp⎡
⎢
⎢
⎣		−
2e−U1/2
U1/2+4
⎤
⎥
⎥
⎦		⎡
⎢
⎢
⎣		1+
8−2U−1/2−U−1+U−3/2
(4+U−1/2)3
 e−U−2U1/2+O(e−2U)⎤
⎥
⎥
⎦		, 
(3.b)

Cette équation a le numéro 3.c, celles d’au dessus les numéro 3.a et 3.b.

  1/t2(y0(x+g)) = y0(x+log(1+y0−2e−y0))
= y0+
e−y0
y02(1+2/y0)
 −
1+4/y0+2/y02
2y04(1+2/y0)3
e−2y0+O(e−3y0).

(3.c)

Part II
Other tests

4 Zéro

x2+y2 = z2 (4.a)
x2+y2 = z2
x2+y2 = z2 (4.b)
x2+y2 = z2
x2+y2 = z2 (oups)

Au desus 4.a, 4.b et oups. Après 4.b

5 Deux

x2+y2 = z2(5.a)
x3+y3 = z3(5.b)
x4+y4 = z4(5.c)
x5+y5 = z5(5.d)
x6+y6 = z6(5.e)
x7+y7 = z7 (5.f)

6 Un

x2+y2 = z2  (6.a)
x3+y3 = z3
x4+y4 = z4  (*)
x5+y5 = z5 *
x6+y6 = z6 6.a′
x7+y7 = z7 (6.b)

Ben mon vieux pour avoir l’équation *.

7 Trois

x2+y2 = z2(+)
x3+y3 = z3
x4+y4 = z4
x5+y5 = z5
x6+y6 = z6
x7+y7 = z7

8 Quatre

x2+y2 = z2(8.a)
x3+y3 = z3(8.b)
x4+y4 = z4(8.c)
x5+y5 = z5(8.d)
x6+y6 = z6(8.e)
x7+y7 = z7 (8.f)

Maintenant : 8.f.

9 Cinq

x2+y2= z2   x3+y3= z3       (a)
x4+y4= z4   x5+y5= z5        
x6+y6= z6   x7+y7= z7       (9.a)

Au dessus ya l’équation (a).

10 Multline

x+1 = y+2

(10.a)

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